Problem: Solve for $x$ : $ 7|x + 8| + 5 = 1|x + 8| + 6 $
Subtract $ {1|x + 8|} $ from both sides: $ \begin{eqnarray} 7|x + 8| + 5 &=& 1|x + 8| + 6 \\ \\ { - 1|x + 8|} && { - 1|x + 8|} \\ \\ 6|x + 8| + 5 &=& 6 \end{eqnarray} $ Subtract ${5}$ from both sides: $ \begin{eqnarray} 6|x + 8| + 5 &=& 6 \\ \\ { - 5} &=& { - 5} \\ \\ 6|x + 8| &=& 1 \end{eqnarray} $ Divide both sides by ${6}$ $ \dfrac{6|x + 8|} {{6}} = \dfrac{1} {{6}} $ Simplify: $ |x + 8| = \dfrac{1}{6}$ Because the absolute value of an expression is its distance from zero, it has two solutions, one negative and one positive: $ x + 8 = -\dfrac{1}{6} $ or $ x + 8 = \dfrac{1}{6} $ Solve for the solution where $x + 8$ is negative: $ x + 8 = -\dfrac{1}{6} $ Subtract ${8}$ from both sides: $ \begin{eqnarray} x + 8 &=& -\dfrac{1}{6} \\ \\ {- 8} && {- 8} \\ \\ x &=& -\dfrac{1}{6} - 8 \end{eqnarray} $ Change the ${ - 8}$ to an equivalent fraction with a denominator of $6$ $ x = - \dfrac{1}{6} {- \dfrac{48}{6}} $ $ x = -\dfrac{49}{6} $ Then calculate the solution where $x + 8$ is positive: $ x + 8 = \dfrac{1}{6} $ Subtract ${8}$ from both sides: $ \begin{eqnarray} x + 8 &=& \dfrac{1}{6} \\ \\ {- 8} && {- 8} \\ \\ x &=& \dfrac{1}{6} - 8 \end{eqnarray} $ Change the ${ - 8}$ to an equivalent fraction with a denominator of $6$ $ x = \dfrac{1}{6} {- \dfrac{48}{6}} $ $ x = -\dfrac{47}{6} $ Thus, the correct answer is $x = -\dfrac{49}{6} $ or $x = -\dfrac{47}{6} $.